# Internal energy, Enthalpy, and Heat

Setting a general background for my question:

We know that the equation of state for an ideal gas is given by:

$$pv=RT$$

where:

• $p$ is the absolute pressure of the gas.
• $v$ is the volume per unit mass of the gas.
• $R$ is the gas constant.
• $T$ is the absolute temperature.

We also know that, following Maxwell equations, the internal energy solely depends on the temperature: $$e=e(T)$$

Introducing enthalpy ($h\equiv e+pv$) and using the equation of state for an ideal gas, we can conclude from Maxwell equations that the enthalpy also depends solely on the temperature: $$h=h(T)$$

Question:

The book I'm using to study Thermodynamics of propulsion, gives a word of warning:

Both the internal energy and the enthalpy can change in the complete absence of heat.

We are still considering ideal gases. However, heat is closely related to change in temperature, and both $e$ and $h$ depend solely on $T$.

My question might be considered a bit basic (I'm starting with Thermodynamics), but I can't quite understand the word of warning the authors gave. How is it possible that $h$ and $e$ change in complete absence of heat, if both depend on $T$? Can you give an example?

Trying to solve my own question:

One example I can think of where $e$ would change in absence of heat is when compressing a gas (I don't know if this is true at all). But I don't have a way to visualize the enthalpy.

• If you're looking for a physical interpretation of enthalpy, don't spend too much time looking. Enthalpy is just a convenient function to work with in solving many kinds of thermodynamics problems. In my judgment, it is best just to think of h as a function defined by the equation you gave. If e changes, then, for an ideal gas T must have changed, so h must have changed. h has changed as a result of e changing and also as a result of pv changing. – Chet Miller Mar 6 '16 at 14:23
• Thanks for the clarification @ChesterMiller. I'm also looking for an answer to my question above. – Jose Lopez Garcia Mar 6 '16 at 14:39
• As you yourself noted, if the gas is compressed (or expanded) adiabatically (i.e., "without heat"), it will do work on its surroundings (or vice versa) and its internal energy, enthalpy, and temperature will change. – Chet Miller Mar 6 '16 at 16:00

Don't associate heat too closely with temperature. That's a connection we make in everyday language, but can be dangerous is physics.

Better: don't make that association at all. Connections between the two exist, for example in the equation for specific heat, but relationships such as that only relate two different concepts.

• Does this mean that a gas can exchange heat, $Q$, without changing its own temperature $T$? How? Thanks. (Can't upvote because I don't have enough reputation). – Jose Lopez Garcia Mar 6 '16 at 14:59
• Yes, if at the same time work is done. (For ideal gases only. The situation can be different for real gases. If a phase transition is possible, then $Q$ can be transferred and the temperature remains constant even in the absence of work.) – garyp Mar 6 '16 at 15:01